Theorem 3impexpbicom 1426

Description: 3impexp 1425 with biconditional consequent of antecedent that is commuted in consequent. (Contributed by Alan Sare, 31-Dec-2011.)

Assertion

Ref	Expression
3impexpbicom	⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ 𝜏)) ↔ (𝜑 → (𝜓 → (𝜒 → (𝜏 ↔ 𝜃)))))

Proof of Theorem 3impexpbicom

Step	Hyp	Ref	Expression
1		bicom 139	. . . 4 ⊢ ((𝜃 ↔ 𝜏) ↔ (𝜏 ↔ 𝜃))
2		imbi2 236	. . . . 5 ⊢ (((𝜃 ↔ 𝜏) ↔ (𝜏 ↔ 𝜃)) → (((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ 𝜏)) ↔ ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜏 ↔ 𝜃))))
3	2	biimpcd 158	. . . 4 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ 𝜏)) → (((𝜃 ↔ 𝜏) ↔ (𝜏 ↔ 𝜃)) → ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜏 ↔ 𝜃))))
4	1, 3	mpi 15	. . 3 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ 𝜏)) → ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜏 ↔ 𝜃)))
5	4	3expd 1214	. 2 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ 𝜏)) → (𝜑 → (𝜓 → (𝜒 → (𝜏 ↔ 𝜃)))))
6		3impexp 1425	. . . 4 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜏 ↔ 𝜃)) ↔ (𝜑 → (𝜓 → (𝜒 → (𝜏 ↔ 𝜃)))))
7	6	biimpri 132	. . 3 ⊢ ((𝜑 → (𝜓 → (𝜒 → (𝜏 ↔ 𝜃)))) → ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜏 ↔ 𝜃)))
8	7, 1	syl6ibr 161	. 2 ⊢ ((𝜑 → (𝜓 → (𝜒 → (𝜏 ↔ 𝜃)))) → ((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ 𝜏)))
9	5, 8	impbii 125	1 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → (𝜃 ↔ 𝜏)) ↔ (𝜑 → (𝜓 → (𝜒 → (𝜏 ↔ 𝜃)))))

Syntax hints:   → wi 4   ↔ wb 104   ∧ w3a 968

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107

This theorem depends on definitions:  df-bi 116  df-3an 970

This theorem is referenced by: (None)